Answer
$4x$
Work Step by Step
We find:
$\textbf{F}\times\textbf{G}=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\2x&1&4y\\x&y&-z\end{vmatrix}$
$=\textbf{i}(-z-4y^{2})-\textbf{j}(-2xz-4yx)+\textbf{k}(2xy-x)$
$=(-4y^{2}-z)\textbf{i}+(2xz+4yx)\textbf{j}+(2xy-x)\textbf{k}$
$\nabla\cdot \textbf{F}=\nabla\cdot\langle -4y^{2}-z,2xz+4yx,2xy-x\rangle$
$=\frac{\partial}{\partial x}(-4y^{2}-z)+\frac{\partial}{\partial y}(2xz+4yx)+\frac{\partial}{\partial z}(2xy-x)$
$=0+4x+0=4x$
